如何区分macOS新Core Audio API的输入/输出AudioHardwareControl？

macOS Sequoia中推出了纯Swift版Core Audio API，无需使用不安全指针即可获取默认输出设备及其所有可用控件，但存在无法区分控件属于输入还是输出的问题：

let system = AudioHardwareSystem.shared
guard let defaultOutputDevice = try system.defaultOutputDevice else { return }
let allControls = try defaultOutputDevice.controls
let volumeControls = allControls.filter { (try? $0.classID) == kAudioVolumeControlClassID }
print(volumeControls) // 2个音量控件，包含输入和输出，但无法区分

虽然可以沿用旧方法实现精准获取：

let system = AudioHardwareSystem.shared
guard let defaultOutputDevice = try system.defaultOutputDevice else { return }
let address = AudioObjectPropertyAddress(mSelector: kAudioHardwareServiceDeviceProperty_VirtualMainVolume,
                                         mScope: kAudioDevicePropertyScopeOutput,
                                         mElement: kAudioObjectPropertyElementMain)
guard defaultOutputDevice.hasProperty(address: address) else { return }
let data = try defaultOutputDevice.propertyData(address: address)
let volume = data.withUnsafeBytes { $0.load(as: Float32.self) }

但通过纯Swift API本身即可区分AudioHardwareControl对应的输入/输出类型：

利用AudioHardwareControl的scope属性，该属性返回的AudioObjectPropertyScope值与旧API中的作用域常量直接对应，可精准筛选输入或输出控件：

let system = AudioHardwareSystem.shared
guard let defaultOutputDevice = try system.defaultOutputDevice else { return }
let allControls = try defaultOutputDevice.controls
let volumeControls = allControls.filter { (try? $0.classID) == kAudioVolumeControlClassID }

// 获取输出音量控件
let outputVolumeControls = volumeControls.filter {
    (try? $0.scope) == kAudioDevicePropertyScopeOutput
}

// 获取输入音量控件
let inputVolumeControls = volumeControls.filter {
    (try? $0.scope) == kAudioDevicePropertyScopeInput
}

内容的提问来源于stack exchange，提问作者Alexander Vasenin